THE COLLECTION. BOOK YIII 
43] 
Now, by hypothesis, 
GE: EA = BD : DC, 
whence CA : AE = BO: CD, 
and, if we halve the antecedents, 
AK:AE= HG: CD] 
therefore A K : EK — HG: HD or BII: HD. 
A 
whence, componendo, GE: EK = BD : DH. (1) 
But AF:FB= BD: DC = {BD : DH). {DH: DC) 
— {GE: EK). {DH: DC). (2) 
Now, ELD being a transversal cutting the sides of the 
triangle KHG, we have 
HL : KL = {GE:EK). {DH: (3) 
[This is ‘ Menelaus’s theorem ’; Pappus does not, however, 
quote it, but proves the relation ad hoc in an added lemma by 
drawing CM parallel to DE to meet HK produced in M. The 
proof is easy, for HL . LK = ( HL . LM) _ ( LM . LK) 
= {HD:DC).{GE:EK).] 
It follows from (2) and (3) that 
AF: FB = HL: LK, 
and, since AB is parallel to HK, and AH, BK are straight 
lines meeting in G, FGL is a straight line. 
[This is proved in another easy lemma by reductio ad 
ahsurdum.]